Answers for Data Skill Challenges for every chapter in the book can be found to check your performance and widen your understanding.

1)      RQ: Are Jill’s employees’ December sales different from the regional averages for per employee sales?

H₀: μ = 14

H₁: μ ≠ 14

α = .05

t(55)_crit = ±2.009

t(55) = 1.97, p = .054 (Excel/SPSS) –or– t(55) = 1.97, p > .05 (by hand)

CI₉₅ = [13.98, 16.42]

Conclusion: Retain the null. The difference is not statistically significant. There is insufficient evidence to conclude that Jill’s employees’ December sales are different from the regional averages. If we assume this sample to represent the population, we would expect 95% of sample means to fall between 13.98 and 16.42.

2)      RQ: Are Sue’s taste testers harsher than the national average?

H₀: μ ≥ 3.5

H₁: μ < 3.5

α = .05

z_crit = –1.645

z = –3.35, p = .000 (Excel/SPSS) –or– z = –3.35, p<= .05 (by hand)

CI₉₅ = [2.33, 2.87]

d = –0.75

Conclusion: Reject the null and accept the alternative. The difference is statistically significant. Sue’s taste testers are harsher than the national average. If we assume this sample to represent the population, we would expect 95% of sample means to fall between 2.33 and 2.87. The difference between Sue’s taste testers and the national sample is a medium effect.

3)      RQ: Does Larry’s company have fewer accidents per year than the average for his industry?

H₀: μ ≥ 15.6

H₁: μ < 15.6

α = .05

t(8)_crit = –1.86

t(8) = –9.50, p = .000 (Excel/SPSS) –or– t(8) = –9.50, p < .05 (by hand)

CI₉₅ = [10.37, 12.41]

Conclusion: Reject the null and accept the alternative. The difference is statistically significant. There is sufficient evidence to conclude that Larry’s company had fewer accidents compared with the national average. If we assume this sample to represent the population, we would expect 95% of sample means would fall between −4.62 and −2.59.

4)      RQ: Does Petra’s branch receive more complaints than the company average?

H₀: μ ≤ 24

H₁: μ > 24

α = .05

z_crit = 1.645

z = 1.96, p = .025 (Excel/SPSS) –or– z = 1.96, p<= .05 (by hand)

CI₉₅ = [24, 33.2]

d = 0.88

Conclusion: Reject the null and accept the alternative. The difference is statistically significant. Petra’s branch receives more complaints than the company average. If we assume this sample to represent the population, we would expect 95% of sample means to fall between 24 and 33.2. The difference between Petra’s branch and the company is a large effect.